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en/10.1.7.md
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| + | ### Statement | ||
| + | |||
| + | $10.1.7.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Problem | ||
| + | |||
| + | $10.1.7.$ An electron enters a region of magnetic field of width $l$. The velocity of the electron $v$ is perpendicular to both the magnetic field induction $B$ and the boundaries of the region. At what angle to the boundary of the region will the electron exit the magnetic field? | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + |  | ||
| + | |||
| + | An electron in a magnetic field moves along a circular arc. The radius of this trajectory $R$ is determined from the equality of the Lorentz force and the centripetal force: | ||
| + | $$e v B = \frac{m_e v^2}{R} \implies R = \frac{m_e v}{e B}$$ | ||
| + | |||
| + | Depending on the width of the region $l$, two cases are possible. Let us find the required angle $\beta$ between the velocity vector and the boundary of the region: | ||
| + | |||
| + | 1. Case $l \le R$ (that is, $\frac{eB}{m_e} \le \frac{v}{l}$) | ||
| + | The electron flies through the region and exits through the right boundary. During its motion, its velocity vector turns by an angle $\theta$. From the geometry of the diagram, the sine of this angle is equal to the ratio of the region's width $l$ to the radius $R$: | ||
| + | $$\sin\theta = \frac{l}{R} = \frac{e B l}{m_e v}$$ | ||
| + | |||
| + | The required angle $\beta$ between the direction of exit of the electron and the boundary itself (the vertical line) complements the angle $\theta$ to $90^\circ$: $\beta = \frac{\pi}{2} - \theta$. Consequently: | ||
| + | $$\cos\beta = \sin\theta = \frac{e B l}{m_e v} \implies \beta = \arccos\left(\frac{e B l}{m_e v}\right)$$ | ||
| + | *(If $l = R$, then $\cos\beta = 1$, yielding $\beta = 0$ — the electron exits tangentially, parallel to the boundary).* | ||
| + | |||
| + | 2. Case $l > R$ (that is, $\frac{eB}{m_e} > \frac{v}{l}$) | ||
| + | The electron does not reach the right boundary. It describes a semicircle in the magnetic field and exits back through the left boundary. At the moment of exit, its velocity vector is directed strictly perpendicular to the left boundary. Consequently, the angle with the boundary is: | ||
| + | $$\beta = \frac{\pi}{2}$$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | $$ | ||
| + | \beta = | ||
| + | \begin{cases} | ||
| + | \arccos\left(\frac{e B l}{m_e v}\right), & \text{for } l \le \frac{m_e v}{e B} \\\\ | ||
| + | \frac{\pi}{2}, & \text{for } l > \frac{m_e v}{e B} | ||
| + | \end{cases} | ||
| + | $$ | ||
| + | |||
| + | --- | ||
| + | **Note on the official answer:** | ||
| + | In the answer key of Savchenko's problem book, it is stated: $\sin\alpha = \frac{eBl}{m_e v}$ (for the first case) and $\alpha = \pi$ (for the second case). The letter $\alpha$ in the official answer denotes the "angle of rotation of the velocity vector" relative to its initial direction (that is, the angle between the initial and final velocity). However, upon a literal reading of the question "At what angle to the boundary of the region...", the solution is the angle $\beta$ derived above. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $10.1.7.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Problem | |||
| $10.1.7.$ An electron enters a region of magnetic field of width $l$. The velocity of the electron $v$ is perpendicular to both the magnetic field induction $B$ and the boundaries of the region. At what angle to the boundary of the region will the electron exit the magnetic field? | |||
| ### Solution | |||
|  | |||
| An electron in a magnetic field moves along a circular arc. The radius of this trajectory $R$ is determined from the equality of the Lorentz force and the centripetal force: | |||
| $$e v B = \frac{m_e v^2}{R} \implies R = \frac{m_e v}{e B}$$ | |||
| Depending on the width of the region $l$, two cases are possible. Let us find the required angle $\beta$ between the velocity vector and the boundary of the region: | |||
| 1. Case $l \le R$ (that is, $\frac{eB}{m_e} \le \frac{v}{l}$) | |||
| The electron flies through the region and exits through the right boundary. During its motion, its velocity vector turns by an angle $\theta$. From the geometry of the diagram, the sine of this angle is equal to the ratio of the region's width $l$ to the radius $R$: | |||
| $$\sin\theta = \frac{l}{R} = \frac{e B l}{m_e v}$$ | |||
| The required angle $\beta$ between the direction of exit of the electron and the boundary itself (the vertical line) complements the angle $\theta$ to $90^\circ$: $\beta = \frac{\pi}{2} - \theta$. Consequently: | |||
| $$\cos\beta = \sin\theta = \frac{e B l}{m_e v} \implies \beta = \arccos\left(\frac{e B l}{m_e v}\right)$$ | |||
| *(If $l = R$, then $\cos\beta = 1$, yielding $\beta = 0$ — the electron exits tangentially, parallel to the boundary).* | |||
| 2. Case $l > R$ (that is, $\frac{eB}{m_e} > \frac{v}{l}$) | |||
| The electron does not reach the right boundary. It describes a semicircle in the magnetic field and exits back through the left boundary. At the moment of exit, its velocity vector is directed strictly perpendicular to the left boundary. Consequently, the angle with the boundary is: | |||
| $$\beta = \frac{\pi}{2}$$ | |||
| #### Answer | |||
| $$ | |||
| \beta = | |||
| \begin{cases} | |||
| \arccos\left(\frac{e B l}{m_e v}\right), & \text{for } l \le \frac{m_e v}{e B} \\\\ | |||
| \frac{\pi}{2}, & \text{for } l > \frac{m_e v}{e B} | |||
| \end{cases} | |||
| $$ | |||
| --- | |||
| **Note on the official answer:** | |||
| In the answer key of Savchenko's problem book, it is stated: $\sin\alpha = \frac{eBl}{m_e v}$ (for the first case) and $\alpha = \pi$ (for the second case). The letter $\alpha$ in the official answer denotes the "angle of rotation of the velocity vector" relative to its initial direction (that is, the angle between the initial and final velocity). However, upon a literal reading of the question "At what angle to the boundary of the region...", the solution is the angle $\beta$ derived above. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||